# Simple Maths Problem

February 21, 2010 Leave a comment

Here was a simple maths problem that was in a magazine (‘New Scientist’) a few days ago. I post this because I think it is a fairly elegant solution to one of these problems which would normally be attacked by brute force.

The question is: Solve the following equation, where *a, b *and *c* are positive integers:

The first step is to realise that 0.138 = 69 ÷ 500. Then invert the equation:

Now we can write this as:

As all numbers are positive, . Therefore the fraction that is the second term on the left hand side must be less than 1. Hence **c = 7**.

Subtract 7 from both sides and multiplying through, we now have that . Factorising, we can write that as

Now 17 is prime, and so one or the other factors must be 1. But it cannot be *b*, as otherwise 17 | 69, which cannot be true as 69 is 3 × 23. Thus we have 17*a* – 69 = 1, and so ** a = 4**, and

**. This must be the only solution as all other cases were ruled out.**

*b*= 17This can be generalised to other numbers than 0.138, of course, but it doesn’t seem too interesting.